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2. ### Collection of Lame Jokes

Being keen in Esperanto I enquired into the opportunity to participate in an exchange program with someone from Esperantoland, because I thought it would an enriching experience.
3. ### Computing path integral with real and Grassmann variables

Performing the ##x## integration exactly for ##Z[w]## could very well not be possible. You may very well not be able to obtain the exact value even if you were just considering this integral: \begin{align*} \frac{1}{(2 \pi)^{n/2}} \int d^nx \exp \left( - \frac{1}{2} w_i (x) w_i (x) \right)...
4. ### POTW An Integral with Fractional Part

@DrClaude you do a shift in the summation variable ##n## followed by extending the sum: ##\sum_{n=1}^\infty n \frac{1}{2 (n+1)^2} = \sum_{n=2}^\infty (n-1) \frac{1}{2 n^2} = \sum_{n=1}^\infty (n-1) \frac{1}{2 n^2}##.

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7. ### POTW Hermitian Matrices

If we use ##\text{rank} (H) = \text{rank} (H^\dagger)## we can simplify a bit my first proof, post #2:
8. ### POTW Hermitian Matrices

No, your idea worked I think. There are no ##u## such that ##H u \not=0## and ##H^\dagger H u =0##, so ##\ker (H) = \ker(H^\dagger H)##. It follows then from ##{H^\dagger}^2 = H^\dagger H## that ##\ker (H) = \ker ({H^\dagger}^2)##. The second question is are there ##u## such that ##H^\dagger u...
9. ### POTW Hermitian Matrices

It would be good there was an easy way of proving ##\ker (H) = \ker (H^\dagger)## as this was the bulk of my solution in post #2. If I understand you you are saying if ##\dim \ker (H) = \dim \ker (H^\dagger)## then does ##{H^\dagger}^2 \ker (H) = H^\dagger H \ker (H)## imply ##\ker (H) = \ker...

Yes.
11. ### POTW Hermitian Matrices

You use this in the proof of the spectral theorem for Hermitian matrices.
12. ### POTW Hermitian Matrices

We can simplify my second proof:
13. ### POTW Hermitian Matrices

I get: \begin{align*} \| Hx -H^\dagger x \|^2 = \langle (H H^\dagger - H^\dagger H) x , x \rangle \end{align*} Do we immediately know ##H## is a normal matrix?

19. ### Residue Theorem applied to a keyhole contour

The sum of residues is indeed equal to: \begin{align*} \dfrac{\sin \left( \frac{3 \pi}{8} \right)}{\sin \left( \frac{3 \pi}{4} \right)} \end{align*} You then just substitute ##\sin \left( \frac{3 \pi}{4} \right) = \frac{1}{\sqrt{2}}## into it to obtain the desired result.
20. ### Residue Theorem applied to a keyhole contour

The denominator of the 2nd term is the negative of the denominator of the first term. That allows you to write your expression as a single fraction. And then, yes, you use ##\sin (x) = \frac{e^{i x} - e^{-i x}}{2i}##.
21. ### Residue Theorem applied to a keyhole contour

Do the same thing with the root ##e^{-\frac{3\pi i}{4}}##. Add the two residues, then multiply by ##2 \pi i##.
22. ### Residue Theorem applied to a keyhole contour

The values of ##z_0## seem right. What are they in polar form?
23. ### Residue Theorem applied to a keyhole contour

This is an example:
24. ### I Schwinger-Dyson equations derivation

Write down ##\theta (-t)## in terms of ##\theta (t)##. Drawing the graph of ##\theta (-t)## helps.
25. ### I Limit as a function, not a value

Just do ##e^{p \log x} = 1 + p \log x + \frac{1}{2!} (p \log x)^2 + \cdots##.
26. ### I Limit as a function, not a value

\begin{align*} \lim_{p \rightarrow 0} \dfrac{x^p - 1}{p} = \lim_{p \rightarrow 0} \dfrac{e^{p \log x} - 1}{p} = \log x . \end{align*}
27. ### POTW Ramanujan Sums: Showing $c_n(k)$

Write \begin{align*} f (n) = \sum_{m=1 , (m,n)=1}^n \exp \left\{ 2 \pi i \frac{m}{n} \right\} \end{align*} then \begin{align*} g (n) = \sum_{d|n} \sum_{m=1 , (m,n)=1}^d \exp \left\{ 2 \pi i \frac{m}{d} \right\} = \left\{ \begin{matrix} 1 & n=1 \\ 0 & n>1 \end{matrix} \right. \end{align*} and...
28. ### POTW Definite Integral of a Rational Function

Yes, you get out the additional result: \begin{align*} \int_0^\infty \frac{1-x^2}{1+x^4} dx = 0 \end{align*} You can see this also by doing ##y=1/x## in the following integral: \begin{align*} \int_0^\infty \frac{1}{1+x^4} dx = \int_0^\infty \frac{y^2}{1+y^4} dy \end{align*}

30. ### POTW Definite Integral of a Rational Function

Using ##(2n)!! = 2n (2n-2) \cdots 2 = 2^n n!##, \begin{align*} \sum_{n=0}^\infty (-1)^n \frac{(2n-1)!!}{(2n)!!} = \sum_{n=0}^\infty \frac{(-1)^n}{n!} \frac{(2n-1)!!}{2^n} \end{align*} We have \begin{align*} \frac{1}{\sqrt{x+1}} = \sum_{n=0}^\infty \frac{(-1)^n}{n!} \frac{(2n-1)!!}{2^n} x^n...
31. ### POTW Linear Systems of ODE

There are a couple of other methods of solving the problem.
32. ### POTW Linear Systems of ODE

Because you have a second order differential equation the general solution will involve two unknown constants, so you need ##x(0)## and ##x'(0)##. Now, ##x'(0)## was not specified in the question but that does not present a problem because you are given a formula for ##x'(t)##. Yep? Do you know...
33. ### POTW Linear Systems of ODE

Yes, they are functions of ##t##.
34. ### POTW Linear Systems of ODE

In the previous post I wrote down slightly the wrong matrix. I've now written down the correct matrix and recalculated the eigenvalues. By the way, there are other methods of solving the problem!
35. ### Solve the given equation that involves fractional indices

Wolfram gets 14.7618
36. ### Solve the given equation that involves fractional indices

Write ##y = x^{\frac{1}{6}}##.

38. ### A An identity with Bessel functions

I am not contributing something new. I am taking a result given in Watson and arriving at the same condition that @renormalize arrived at. On page 148 of Watson is the formula: \begin{align*} J_\mu (az) J_\nu (bz) & = \dfrac{ (\frac{1}{2} az)^\mu (\frac{1}{2} b z)^\nu}{\Gamma (\nu+1)}...
39. ### A An identity with Bessel functions

It's not exactly "drivng me batty", but let's switch from ##e## to ##x## anyway. Say we have \begin{align*} A J_2 (x) + B Y_2 (x)= \frac{1}{x} \sum_{n=1}^\infty \dfrac{J_n (n x)}{n} \dfrac{J_{n+1} ((n+1) x)}{n+1} \end{align*} We have that ##J_2 (0) = 0## and the RHS is zero at ##x=0## (because...
40. ### A An identity with Bessel functions

I know you are considering ##0 < e < 0##. But can we not use that ##\lim_{e \rightarrow 0} J_2 (e) \rightarrow 0## and that ##Y_2 (e)## is singular for ##e=0##? Wouldn't the sum: \begin{align*} \frac{1}{e} \sum_{i=1}^\infty \dfrac{J_i ( i e)}{i} \dfrac{J_{i+1} ((i+1) e)}{i+1} \end{align*}...

42. ### Have any famous or well known scientists come to this site?

I've just remembered, there's this BBC children's programme called Blue Peter which featured the "science sisters" for a while. I was actually doing my PhD with one of the science sisters at the time! It was so surreal to see her on TV. Blue Peter encouraged you to write in with questions. I...
43. ### Have any famous or well known scientists come to this site?

Appearances can be deceiving, Mr Bean has a master's degree in electrical engineering.
44. ### POTW An Inequality of Sylvester

Supplementary
45. ### Have any famous or well known scientists come to this site?

I had no idea! Which page?
46. ### Have any famous or well known scientists come to this site?

John Baez has been posting here from 2004 to 2021: https://www.physicsforums.com/search/7497600/ unless that is somebody pretending to be John Baez. Francesca Vidotto, who wrote the book "Covariant loop quantum gravity" with Carlo Rovelli...